Question concerning the proof of the equality of $x\sqrt{y}$ and $y\sqrt{\frac{x^2}{y}}$ I want to prove that $x\sqrt{y}=y\sqrt{\dfrac{x^2}{y}}$; I've proven it to myself via calculator (brute forcing it) when $x,y>0$, and this is my proof:
$$\begin{align*}
x\sqrt{y} &= y\sqrt{\dfrac{x^2}{y}}&&\text{Conjecture}\\
          &= y\dfrac{\sqrt{x^2}}{\sqrt{y}}&&\text{By Theorem 2.2}\\
          &= y\dfrac{x}{\sqrt{y}}&&\because\sqrt{x^2}=x\\
          &= \dfrac{xy}{\sqrt{y}}&&\text{Simplification}\\
x\sqrt{y}\cdot\dfrac{1}{x} &= \dfrac{xy}{\sqrt{y}}\cdot\dfrac{1}{x}&&\text{Multiplication}\\
\sqrt{y}  &= \dfrac{y}{\sqrt{y}}&&\blacksquare
\end{align*}$$
Questions


*

*In a proof, does it suffice to end with a truism as a result of the conjecture to prove the conjecture?

*Is my proof correct (the steps I took to get there)?

 A: You didn't write it properly. I believe you meant to write the following.
$$\color{blue}{x\sqrt{y} = y\sqrt{\dfrac{x^2}{y}}\tag{Conjecture}}$$
\begin{align*}
          x\sqrt{y} = y\sqrt{\dfrac{x^2}{y}}&\iff x\sqrt y= y\dfrac{\sqrt{x^2}}{\sqrt{y}}&&\text{By} \sqrt{\frac{x}{y}}\text{'s definition}\\
          &\iff x\sqrt y= y\dfrac{x}{\sqrt{y}}&&\because\sqrt{x^2}=x\\
          &\iff x\sqrt y= \dfrac{xy}{\sqrt{y}}&&\text{Simplification}\\
&\iff x\sqrt{y}\cdot\dfrac{1}{x} = \dfrac{xy}{\sqrt{y}}\cdot\dfrac{1}{x}&&\text{Multiplication}\\
&\iff \sqrt{y}  = \dfrac{y}{\sqrt{y}}&&\blacksquare
\end{align*}
The proof itself is correct, but, as it has been pointed out in the comments, the first justification is wrong. this answers the second question.
Answering the first question, yes, that is enough, you've proved that the conjecture is true if, and only if, something which is always true, is true, therefore the conjecture is always true.
You only need the direction going from bottom to top, though, so you can replace all the $\iff$'s by $\Longleftarrow$'s to get 
$$\sqrt y=\dfrac y{\sqrt y}\implies x\sqrt y=y\sqrt{\dfrac{x^2}{y}}.$$
Since the above statement is true (because of the given proof) and because the antecedent of the  conditional statement is true, then the consequent can't be false.
